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KIM [24]
1 year ago
9

I) Hence solve cosec 3y ÷ cot 3y + tan 3y = 0.5 for 0

Mathematics
1 answer:
nata0808 [166]1 year ago
4 0

The value of y is π/9 or 7π/9 for the range 0 ≤ y ≤ π.

cosec 3y /( cot 3y + tan 3y) = 0.5

Trigonometry formula is:

cosec Ф = 1/ sin Ф

cot Ф = cos Ф/ sinФ

tan Ф = sinФ/ cos Ф

\frac{\frac{1}{sin 3y} }{\frac{cos 3y}{sin 3y} +\frac{sin 3y}{cos 3y}  }  = 0.5

\frac{\frac{1}{sin 3y} }{\frac{cos 3y^{2}  + sin 3y^{2} }{sin 3y cos 3y} } = 0.5

sin² 3y + cos ² 3y = 1

1/ sin 3y × sin 3y cos 3y = 0.5

cos 3y = 0.5 = 1/2

cos π/ 3 = 1/2

For 0≤ y ≤π

0 ≤3y ≤ 3π

So 3y = 2kπ + π/3 , k = 0, 1

y = π/ 9 or 7π/ 9

Therefore the value of y is π/9 and 7π/ 9 for the given range.

To know more about trigonometry refer to the link given below:

brainly.com/question/13729598

#SPJ9

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Solve 7x – 9 = 28 + 4(x – 1).
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4 years ago
A) One of the roots of the equation 10x^2−33x+c=0 is 5.3. Find the other root and the coefficient c.
Aleks [24]

Answer:

a) The other root is -2

   The coefficient c = -106

b) q = 35

c) c = -8.75

Step-by-step explanation:

* Lets study the general form of the quadratic equation

* ax² + bx + c = 0

- Their roots are x1 and and x2

- The sum of them = -b/a ⇒ x1 + x2 = -b/a

- The product of them = c/a ⇒ (x1)(x2) = c/a

a) * Assume that the roots of the equation 10x² - 33x + c = 0

     are m and n

∵ m + n = -b/a

∵ a = 10 and b = -33

∴ m + n = -(-33)/10 = 3.3

∵ m = 5.3

∴ 5.3 + n = 3.3 ⇒ n = 3.3 - 5.5 = -2

∴ n = -2

* The other root is -2

∵ m × n = c/a

∵ m = 5.3 , n = -2 , a = 10

∴ (5.3)(-2) = c/10

∴ -10.6 = c/10 ⇒ Multiply both sides by 10

∴ c = -106

* The coefficient c = -106

b) * Assume that the roots of the equation x² - 12x + q = 0

     are m and n

∵ The difference between the roots is 2

∴ m - n = 2 ⇒ (1)

∵ From the equation m + n = -b/a

∵ a = 1 , b = -12

∴ m + n = -(-12)/1 = 12

∴ m + n = 12 ⇒ (2)

* Lets solve the two equation

- Add the two equation to eliminate n

∴ 2m = 14 ⇒ divide both sides by 2

∴ m = 7

* Substitute the value of m in (1) or (2)

∴ 7 - n = 2 ⇒ 7 - 2 = n ⇒ 5 = n

∴ n = 5

∵ mn = c/a

∵ c = q , a = 1

∴ mn = q/1 = q

∴ q = 7 × 5 = 35

* q = 35

c) * Assume that the roots of the equation x² + x + c = 0

     are m and n

∵ The difference between the roots is 6

∴ m - n = 6 ⇒ (1)

∵ From the equation m + n = -b/a

∵ a = 1 , b = 1

∴ m + n = -(1)/1 = -1

∴ m + n = -1 ⇒ (2)

* Lets solve the two equation

- Add the two equation to eliminate n

∴ 2m = 5 ⇒ divide both sides by 2

∴ m = 2.5

* Substitute the value of m in (1) or (2)

∴ 2.5 + n = -1 ⇒ n = -1 - 2.5 = -3.5

∴ n = -3.5

∵ mn = c/a

∵ c = c , a = 1

∴ mn = c/1 = c

∴ c = 2.5 × (-3.5) = -8.75

* c = -8.75

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